Question

9. Calculate the standard enthalpy of
formation of CH,OH(T) from the following
data
LCH,OH()+ 0,69
)
CO (9)+ 2H2O(l),
AH = -726 kJ mol-1
ii. C (Graphite) + 0,(9) CO2(g),
A Hº = -393 kJ mol-
iii. H.(g) + 2 02(9) H0(1),
AH = -286 kJ mol-
Ans. : (-239 kJ moll)​

Answer 1

Answer:

On reversing the first equation we get,

CO

2

(g)+2H

2

O(l)→CH

3

OH(l)+

2

3

O

2

(g); Δ

r

H

θ

=+726kJmol

−1

(on reversing the reaction, the sign of ΔH also reverses)

The second equation remains as such,

C

(graphite)

+O

2

(g)→CO

2

(g);Δ

c

H

θ

=−393kJmol

−1

On multiplying equation three by 2 we get,

2H

2

(g)+O

2

(g)→2H

2

O(l);2Δ

f

H

θ

=2×−286kJmol

−1

=−572kJmol

−1

On adding these three equations we get,

C

(graphite)

+2H

2

(g)+

2

1

O

2

(g)→CH

3

OH(l)

Enthalpy of formation of CH

3

OH=Δ

f

H=Δ

r

H

θ

+Δ

c

H

θ

+2Δ

f

H

θ

Δ

f

H=+726−393−572kJmol

−1

=−239kJmol

−1

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